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16. The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

A. 25% increase B. 50% increase C. 50% decrease D. 75% decrease

Ans. B

Solution:

Let original length = x and original breadth = y. Original area = xy. New length =x/2 New breadth = 3y. New area=[(x/2)*3y]=(3/2)xy Increase % =[(1/2)xy*(1/xy)*100]%=50%

17. The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

A. 1520 m2 B. 2420 m2 C. 2480 m2 D. 2520 m2

Ans. D

Solution:

We have: (l – b) = 23 and 2(l + b) = 206 or (l + b) = 103. Solving the two equations, we get: l = 63 and b = 40. Area = (l x b) = (63 x 40) m2 = 2520 m2.

18. In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years age which is 48. What is the total number of students in the school?

A. 72 B. 80 C. 120 D. 100

Ans. D

19. Point out the error in the program?

#include

int main()

{

struct emp

{

char name[20];

float sal;

};

struct emp e[10];

int i;

for(i=0; i<=9; i++)

scanf(“%s %f”, e[i].name, &e[i].sal);

return 0;

}

A. Error: invalid structure member

B. Error: Floating point formats not linked

C. No error

D. None of above

Ans. B

Solution:

At run time it will show an error then program will be terminated. Sample output: Turbo C (Windwos) Sample 12.123 scanf : floating point formats not linked Abnormal program termination

20. Point out the error in the program?

#include

int main()

{

struct emp

{

char name[20];

float sal;

};

struct emp e[10];

int i;

for(i=0; i<=9; i++)

scanf(“%s %f”, e[i].name, &e[i].sal);

return 0;

}

A. Error: invalid structure member

B. Error: Floating point formats not linked

C. No error

D. None of above

Ans. B

Solution:

At run time it will show an error then program will be terminated. Sample output: Turbo C (Windwos) c:>myprogram Sample 12.123 scanf : floating point formats not linked Abnormal program termination

21. What will be the output of the program?

#include

int main()

{

char p[] = “%dn”;

p[1] = ‘c’;

printf(p, 65);

return 0;

}

A. A

B. a

C. c

D. 65

Ans. A

Solution:

Step 1: char p[] = “%dn”; The variable p is declared as an array of characters and initialized with string “%d” Step 2: p[1] = ‘c’; Here, we overwrite the second element of array p by ‘c’. So array p becomes “%c”. Step 3: printf(p, 65); becomes printf(“%c”, 65); Therefore it prints the ASCII value of 65. The output is ‘A’.

22. What will be the output of the program?

#include

#include

int main()

{

char str1[20] = “Hello”, str2[20] = ” World”;

printf(“%sn”, strcpy(str2, strcat(str1, str2)));

return 0;

}

A. Hello

B. World

C. Hello World

D. WorldHello

Ans. C

Solution:

Step 1: char str1[20] = “Hello”, str2[20] = ” World”; The variable str1and str2 is declared as an array of characters and initialized with value “Hello” and ” World” respectively. Step 2: printf(“%sn”, strcpy(str2, strcat(str1, str2))); => strcat(str1, str2)) it append the string str2 to str1. The result will be stored in str1. Therefore str1 contains “Hello World”. => strcpy(str2, “Hello World”) it copies the “Hello World” to the variablestr2. Hence it prints “Hello World”.

23. Point out the error in the program

#include

int main()

{

int i;

#if A

printf(“Enter any number:”);

scanf(“%d”, &i);

#elif B

printf(“The number is odd”);

return 0;

}

A. Error: unexpected end of file because there is no matching #endif

B. The number is odd

C. Garbage values

D. None of above

Ans. A

Solution:

The conditional macro #if must have an #endif. In this program there is no#endif statement written.

24. Point out the error in the program

#include

#define SI(p, n, r) float si; si=p*n*r/100;

int main()

{

float p=2500, r=3.5;

int n=3;

SI(p, n, r);

SI(1500, 2, 2.5);

return 0;

}

A. 26250.00 7500.00

B. Nothing will print

C. Error: Multiple declaration of si

D. Garbage values

Ans. C

Solution:

The macro #define SI(p, n, r) float si; si=p*n*r/100; contains the error. To remove this error, we have to modify this macro to #define SI(p,n,r) p*n*r/100

25. What will be the output of the program?

#include

int main()

{

float d=2.25;

printf(“%e,”, d);

printf(“%f,”, d);

printf(“%g,”, d);

printf(“%lf”, d);

return 0;

}

A. 2.2, 2.50, 2.50, 2.5

B. 2.2e, 2.25f, 2.00, 2.25

C. 2.250000e+000, 2.250000, 2.25, 2.250000

D. error

Ans. C

Solution:

printf(“%e,”, d); Here ‘%e’ specifies the “Scientific Notation” format. So, it prints the 2.25 as 2.250000e+000. printf(“%f,”, d); Here ‘%f’ specifies the “Decimal Floating Point” format. So, it prints the 2.25 as 2.250000. printf(“%g,”, d); Here ‘%g’ “Use the shorter of %e or %f”. So, it prints the 2.25 as 2.25. printf(“%lf,”, d); Here ‘%lf’ specifies the “Long Double” format. So, it prints the 2.25 as 2.250000.

26. What will be the output of the program?

#include

#include

int main()

{

float n=1.54;

printf(“%f, %fn”, ceil(n), floor(n));

return 0;

}

A. 2.000000, 1.000000

B. 1.500000, 1.500000

C. 1.550000, 2.000000

D. 1.000000, 2.000000

Ans. A

Solution:

ceil(x) round up the given value. It finds the smallest integer not < x. floor(x) round down the given value. It finds the smallest integer not > x. printf(“%f, %fn”, ceil(n), floor(n)); In this line ceil(1.54) round up the 1.54 to 2 and floor(1.54) round down the 1.54 to 1. In the printf(“%f, %fn”, ceil(n), floor(n)); statement, the format specifier “%f %f” tells output to be float value. Hence it prints 2.000000 and 1.000000.

27. What will be the output of the program?

#include

int main()

{

float a=0.7;

if(a < 0.7f)

printf(“Cn”);

else

printf(“C++n”);

return 0;

}

A. C

B. C++

C. Compiler error

D. Non of above

Ans. B

Solution:

if(a < 0.7f) here a is a float variable and 0.7f is a float constant. The float variable a is not less than 0.7f float constant. But both are equal. Hence the ifcondition is failed and it goes to else it prints ‘C++’

28. What will be the output of the program?

#include

int main()

{

float f=43.20;

printf(“%e, “, f);

printf(“%f, “, f);

printf(“%g”, f);

return 0;

}

A. 4.320000e+01, 43.200001, 43.2

B. 4.3, 43.22, 43.21

C. 4.3e, 43.20f, 43.00

D. Error

Ans. A

Solution:

printf(“%e, “, f); Here ‘%e’ specifies the “Scientific Notation” format. So, it prints the 43.20 as 4.320000e+01. printf(“%f, “, f); Here ‘%f’ specifies the “Decimal Floating Point” format. So, it prints the 43.20 as 43.200001. printf(“%g, “, f); Here ‘%g’ “Use the shorter of %e or %f”. So, it prints the 43.20 as 43.2.

29. If the size of an integer is 4 bytes, What will be the output of the program?

#include

#include

int main()

{

printf(“%dn”, strlen(“123456”));

return 0;

}

A. 6

B. 12

C. 7

D. 2 Ans. A

Solution:

The function strlen returns the number of characters int the given string. Therefore, strlen(“123456”) contains 6 characters. The output of the program is “6”.

30. What will be the output of the program?

#include

int main()

{

int arr[5], i=0;

while(i<5)

arr[i]=++i;

for(i=0; i<5; i++)

printf(“%d, “, arr[i]);

return 0;

}



A. 1, 2, 3, 4, 5,

B. Garbage value, 1, 2, 3, 4,

C. 0, 1, 2, 3, 4,

D. 2, 3, 4, 5, 6,

Ans. B

Solution:

Since C is a compiler dependent language, it may give different outputs at different platforms. We have given the TurboC Compiler (Windows) output. Please try the above programs in Windows (Turbo-C Compiler) and Linux (GCC Compiler), you will understand the difference better.

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